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Testing seed mixtures

Testing seed mixtures. Júlia Barabás PhD (Hungary). Introduction In year 2000 two questions were addressed by the Purity Committee to the Statistics Committee. 1. How could we compare the test result of a sample against the labelled proportion of seed mixture-lot?

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Testing seed mixtures

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  1. Testing seed mixtures Júlia Barabás PhD (Hungary)

  2. IntroductionIn year 2000 two questions were addressed by the Purity Committee to the Statistics Committee. 1. How could we compare the test result of a sample against the labelled proportion of seed mixture-lot? (= does the lot has the labelled proportion?) 2. How could we compare two test results from a seed mixture-lot? (= we have made a test on our submitted sample but we or another lab had to perform another test, “are the two results compatible?” )

  3. Suggested methods and their mathematical background Subgroup to discuss the possibilities by e_mail (leadership J. Barabás). After about 6 circulars and different contributions, the mathematical background was chosen and a software was prepared to address 4 different cases

  4. case 1 a when we want to compare the labelled proportion with one test result of the sample on the basis of the count of the number of seeds of different components. In this case we could use a well known statistical test the Chi-squared test for goodness of fit. It measures the difference between the expected and the experienced frequencies.

  5. n=number of seeds tested Pi0=number of seeds of a component

  6. case 1 b when we want to compare the labelled proportion with one test result of the sample and on the basis of the weight of the different components. (we need to know or measure the 1000 seed-weights of the components). We transform the original formula and suggest to apply in this case as follows: Ho and H1and rejection region are the same as in the first case. Ho =all frequencies are as expected H1= at least one of the frequencies is different from expected

  7. Pi0 = the labelled proportion % of the i-th component

  8. Case 2 a when we want to compare two test results from the same seed mixture lot on the basis of the count of the number of seeds of different components in both tests. The base statistic is a homogeneity Chi-squared test. Ho: F1(x)= F2(x), that is the two distributions are the same, H1 : F1(x) not = F2(x) that is the two distributions are not equal.

  9. The test statistic: where h is the seed-number of the i-th i component in the first test and l is the i seed-number of the i-th component in the second test, and n n the total seed- 1 , 2 numbers in each test

  10. 2 b case when we want to compare two test results from the same seed mixture lot on the basis of the weightof different components in both tests. The base statistic is the same but we need to transform the original formula.

  11. Where vi is the weight of the i-th component in the first test and wi is the weight of the i-th component in the second test, and mi the 1000 seed weight. Ho and H1and rejection region are the same as in case 2/a.

  12. Excel spreadsheet to compute tests in seed mixtures • In order to allow the practical use in the many possible circumstances • When work will be validated by Statistic and Purity committee, it will be available through ISTA

  13. Case where seed mixture composition is checked by numbers of seeds

  14. Case where seed mixture composition is checked by weight of seeds

  15. Another question : Which number of seeds shall I expect for a given labelled % and number of seeds

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